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Anderson Video - Spinning Room Amusement Park Ride

Professor Anderson
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>> So you've probably been on this ride before. They used to have it at Magic Mountain, that's where I went on it. But the idea is the following. You are going to stand inside this room. And the room is going to start spinning. And, as you go around in a circle, the floor starts to drop out from underneath you. And, as this thing spins around and the floor drops out underneath you, you don't fall with the floor, okay? So, let's ask you a question. If I'm going to draw the free body diagram, I need to know how many forces are acting on the person. Come on in, guys, grab a seat. There's some extra chairs in the back if you need. How many forces are acting on the person? And I'm going to ask you guys here in class, and I will ask the people at home to chime in on the Blackboard Collaborate. I'll give them some options. And let's say that we've got a few options to consider. And we'll just say, 1, 2, 3, 4 or 5. How many forces are acting on the person? Take a second to discuss it with your groups. People at home, chime in on the chatroom and then give us a vote. Back table, what'd you guys think? >> Four. >> Four? Another one for 4. And what about table -- the second table over here, what'd you guys think? >> Four. >> Also 4? Okay and the last table? >> Four. >> Four. All right, let's see what the people at home decided. Okay, most people at home said it should be 3, but the second most popular was 4. So, let's think about this problem for a second. This is, of course, the side view of this spinning room. So, if I put the free body diagram right here, that person becomes a dot. What are the forces that are acting on the person? What's the first one we always draw? >> Gravity. >> Gravity. Which way is gravity. >> Down, okay. What else do we have? Okay, somebody said friction. Where is friction? >> Behind the wall. >> Okay, friction between the person and the wall. Which direction should I draw friction on here? >> Up. >> Up? Okay. That sounds like a good idea. We need something to combat gravity, right? If that person is not going to fall, if they're stuck to the wall, you can't just have MG all by itself. You've got to have something else holding them up. Is that it? >> Normal force? Which way is the normal force? Inwards. The wall is pushing on the back of the person, and that normal force is, in fact, what keeps them moving in a circle. All right, anything else? Those of you that said 4, what was the fourth one that you were envisioning? Maybe something like this? What would that be? Okay. Centrifugal force? Or Centripetal force? Remember, when we talked about centrifugal force, we said it is a fictional force. It is something that you perceive as going to the right. That person perceives that there's a force going to the right because that motion is around to the left. So, that's not a real force. And centripetal is just a description for the forces that keep you moving in the circle. So is this it? Yeah, this is it. Correct answer is C, 3 forces. There's only 3 forces acting on that person. So, let's see if we can now analyze the kinematics for this problem. Okay? If we have some radius here, R, can we figure out how these things relate to each other? And, in fact, let's figure out the question, what is the slowest, V, such that you don't slip? In other words, you're stuck to the wall, what's the slowest that you can rotate this ride and still have that person stick to the wall? Okay. Anybody been on this ride? You guys ever been on one of these rides? Yeah? It's just, it's a fantastic way to get nauseous. It really will just make you sick almost instantly, especially after 30. For some reason, you just can't do spinny rides anymore. It's really annoying. Okay, good thing I'm only 29. Okay, what is the slowest V that we could possibly use, and how do we figure this out? Well, we've got a picture, we've got a free body diagram. We want to go to Newton's Second Law now. And, specifically, we're worried about Newton's Second in the radial direction and in the vertical direction. The radial direction is not too bad, there's only 1 force. And we know that the sum of all the radial forces has to add up to MB squared over R. Okay? There's only 1, it's the normal force. So we get normal force is equal to MB squared over R. That's one of our equations that we're going to use. For the vertical forces, we, of course, have two forces. We have the frictional force holding us up, we have MG trying to pull us down, and so we get F sub S minus MG equals what? What's the acceleration in the vertical direction? Zero, right? If you're moving around in this horizontal circle, you're not going up or down. And so we just get F sub S equals MG. So, if we want to figure out the slowest velocity that you can go, the slowest speed of rotation, how do we do this? It doesn't look like we have quite enough information yet. Because we have the normal force here, we've got frictional force there. But we have to relate the two. And we sort of know what the answer is, right? Static friction is the following, F sub S is less than, or equal to, Mu S times the normal force. What we want to do is we want to go with the slowest V. So, do you want the wall to be sticky or do you want the wall to be slippery? Sticky, right? You want to be able to stick to the wall so actually you can go at a slower speed. And that means that we are looking at the maximum that static friction can be. S F max which is just Mu S times N. Okay, we want to take advantage of the maximum amount of static friction that we could. Okay, so it looks like we can probably put this stuff together now. Let's see how we do this. We've got an equation here that looks like we could solve it for V. Right? We've got normal force is equal to MV squared over R. Right here, we can put what the normal force is, so this is F S max divided by Mu sub S. Okay, we just took that and plugged it in right there. And we know what F S is. F S is just MG. And so, when I take this over here, we get MG over Mu S equals MB squared over R. All right, hopefully, that was kind of clear. But, if not, you can review the videotape. All right and, so now, we can solve this for V. So, if we're going to solve this for V, look what happens. The M cancels out on both sides. And, if I multiply across by R, I get V is equal to G times an R divided by Mu S and then I have to take the square root. So, let's take a look at this answer, and see if it makes sense. First off, the mass fell out. There's no mass dependence here. Does that make sense? Yeah, it makes sense. Why does it make sense? Because, when you have an amusement park ride, you don't want to have to worry about the mass of the rider being able to stick to the wall or not, right? Likewise, the units look like they work out, right? We've got G which is meters per second squared. We're going to multiply by R which is meters. So, we get meters squared per second squared. When I take the square root of that, it looks like we're going to get meters per second. So the units work out. And, the limits look pretty good. Namely, if Mu S gets big, what happens to the velocity? It gets small. In other words, you can spin much more slowly. The whole room can spin slower which makes sense, right? We talked about this initially. If it's a very slippery wall, that friction's not going to hold you up very well, so you have to spin around much faster. But if it's a very sticky wall, where Mu S gets big, then you can spin the room around much more slowly. Okay? So that looks like a good answer, and you can plug in numbers if you like for your situation, depending on the givens.